Intersecting Line Against Cylinder

Consider a cylinder with radius $r$ and length $\ell$ whose axis contains points $\mathbf{q}_0$ and $\mathbf{q}_1$ . Let $\mathbf{L}(t) = \mathbf{p}_0 + t (\mathbf{p}_1 - \mathbf{p}_0)$ be a line passing through points $\mathbf{p}_0$ and $\mathbf{p}_1$ , and $\mathbf{c}_L = \mathbf{p}_0 + t_c (\mathbf{p}_1 - \mathbf{p}_0)$ be the point in $\mathbf{L}$ closest to the cylinder axis and $\mathbf{c}_c$ the point in the cylinder axis closest to $\mathbf{L}$ . If the distance $f$ between the closest points is smaller than the cylinder radius, a value $\delta$ can be added to and subtracted from $t_c$ to find the intersection points $\mathbf{i}_0 = \mathbf{L}(t_c - \delta)$ and $\mathbf{i}_1 = \mathbf{L}(t_c + \delta)$ on the surface of the cylinder.

Cylinder line intersection

The line and the vector orthogonal to the line and the cylinder axis (i.e. their cross product) define a plane which cuts the cylinder, defining an ellipse as their intersection. The length of the semi-minor axis of this ellipse is $r$ , and the length of the semi-major axis is $s$ , which is unknown.

The the value of $s$ is inversely proportional to the sine of the angle between the cylinder axis and $\mathbf{L}$ . At 90 degrees, $s = r$ . As the angle decreases, $s$ increases.

$r = s \sin \theta$

$r^2 = s^2 \sin^2 \theta$

$r^2 = s^2 (1 - \cos^2 \theta)$

$s^2 = \frac{r^2 } {1 - \cos^2 \theta}$

Let $\mathbf{d} = \mathbf{p}_1 - \mathbf{p}_0$ and $\mathbf{e} = \mathbf{q}_1 - \mathbf{q}_0$ . Using the dot product:

$\mathbf{d} \cdot \mathbf{e} = \| \mathbf{d} \| \| \mathbf{e} \| \cos \theta$

$\left( \mathbf{d} \cdot \mathbf{e} \right)^2 = (\mathbf{d} \cdot \mathbf{d}) \, (\mathbf{e} \cdot \mathbf{e}) \cos^2 \theta$

$\cos^2 \theta = \frac{\left( \mathbf{d} \cdot \mathbf{e} \right)^2}{ (\mathbf{d} \cdot \mathbf{d}) \, (\mathbf{e} \cdot \mathbf{e})}$

Continuing on the development of $s$ :

$s^2 = \frac{r^2} {1 - \frac{\left( \mathbf{d} \cdot \mathbf{e} \right)^2}{(\mathbf{d} \cdot \mathbf{d}) \, (\mathbf{e} \cdot \mathbf{e})}}$

$s^2 = \frac{r^2 (\mathbf{d} \cdot \mathbf{d}) \, (\mathbf{e} \cdot \mathbf{e})} {(\mathbf{d} \cdot \mathbf{d}) \, (\mathbf{e} \cdot \mathbf{e}) - \left( \mathbf{d} \cdot \mathbf{e} \right)^2}$

Our ellipse equation is

$\frac{g^2}{s^2} + \frac{f^2}{r^2} = 1$

where $f$ is the distance between $\mathbf{L}$ and the cylinder axis and $g$ is how much the closest point on $\mathbf{L}$ must be offset to lie on the surface of the cylinder. Solving for $g$ :

$g^2 = s^2 \left(1 - \frac{f^2}{r^2} \right)$

$= \frac{r^2 (\mathbf{d} \cdot \mathbf{d}) \, (\mathbf{e} \cdot \mathbf{e})} {(\mathbf{d} \cdot \mathbf{d}) \, (\mathbf{e} \cdot \mathbf{e}) - \left( \mathbf{d} \cdot \mathbf{e} \right)^2} \frac{r^2 - f^2}{r^2}$

$= \frac{(\mathbf{d} \cdot \mathbf{d}) \, (\mathbf{e} \cdot \mathbf{e}) (r^2 - f^2)} {(\mathbf{d} \cdot \mathbf{d}) \, (\mathbf{e} \cdot \mathbf{e}) - \left( \mathbf{d} \cdot \mathbf{e} \right)^2}$

The value of $g$ is a measure of length, while $\delta$ is a fraction of the length of $\mathbf{d}$ . To obtain $\delta$ , $g$ must be divided by the length of $\mathbf{d}$ .

$\delta = \frac{g}{\| \mathbf{d} \|}$

$\delta^2 = \frac{g^2}{\mathbf{d} \cdot \mathbf{d}}$

$\delta^2 = \frac{(\mathbf{d} \cdot \mathbf{d}) \, (\mathbf{e} \cdot \mathbf{e}) (r^2 - f^2)} {(\mathbf{d} \cdot \mathbf{d}) \, (\mathbf{e} \cdot \mathbf{e}) - \left( \mathbf{d} \cdot \mathbf{e} \right)^2} \frac{1}{(\mathbf{d} \cdot \mathbf{d})}$

$\delta^2 = \frac{(\mathbf{e} \cdot \mathbf{e}) (r^2 - f^2)} {(\mathbf{d} \cdot \mathbf{d}) \, (\mathbf{e} \cdot \mathbf{e}) - \left( \mathbf{d} \cdot \mathbf{e} \right)^2}$

Now the first and second intersection points $\mathbf{i}_0$ and $\mathbf{i}_1$ can be calculated using $\delta$ .